Binomial Theorem Test

Result

Binomial Theorem Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

In the expansion of $${\left( {3 - \sqrt {\frac{{17}}{4} + 3\sqrt 2 } } \right)^{15}}$$ the 11th term is a

A
Positive integer

B
positive irrational number

C
negative integer

D
negative irrational number

Solution
Question 2
1 / -0

In the expansion of $$\displaystyle \left ( 3 -\sqrt{\dfrac{17}{4} + 3\sqrt{2}} \right )^{15}$$ the $$11^{th}$$ term is a :

A
positive integer

B
positive irrational number

C
negative integer

D
negative irrational number

Solution
Question 3
1 / -0

The coefficient of $${ x }^{ 4 }$$ in the expansion of $${ (1+x+{ x }^{ 2 }+x }^{ 3 })^{ 4 }$$ is

A
$$^{ n }{ C }_{ 4 }$$

B
$$^{ n }{ C }_{ 4 }+^{ n }{ C }_{ 2 }$$

C
$$^{ n }{ C }_{ 4 }+^{ n }{ C }_{ 2 }+^{ n }{ C }_{ 4 }.^{ n }{ C }_{ 2 }$$

D
$$^{ n }{ C }_{ 4 }+^{ n }{ C }_{ 2 }+^{ n }{ C }_{ 1 }-^{ n }{ C }_{ 2 }$$

Solution

$$ {\textbf{Step - 1: Factorise the polynomial }} $$

$$ {\left( {1 + x + {x^2} + {x^3}} \right)^4}\; $$

$$   = {\left[ {1\left( {1 + x} \right) + {x^2}\left( {1 + x} \right)} \right]^4} $$

$$   = {\left( {1 + x} \right)^4}{\left( {1 + {x^2}} \right)^4} $$

$$ {\textbf{Step - 2: Calculation}} $$

$$ {\left( {1 + x + {x^2} + {x^3}} \right)^4} = {\text{ }}{\left( {1 + x} \right)^4}{\left( {1 + {x^2}} \right)^4} $$

$$ {\text{Coefficient of }}{x^{4\;}} $$

$$   \Rightarrow {\;^n}{C_0}^n{C_2}{ + ^{\;n}}{C_2}{.^{\;n}}{C_1}{ + ^n}{C_2}{.^n}{C_4} $$

$$ { = ^n}{C_4}{ + ^n}{C_2}{ + ^n}{C_4}{.^n}{C_2} $$

$$ {{\textbf{ Hence, the correct answer is option C}}{\text{.}}} $$
Question 4
1 / -0

The coefficient of $$x^n$$ in the expansion of $$\frac{1}{{(1 - x)(3 - x)}}$$ is

A
$$\frac{{{3^{n + 1}} - 1}}{{{{2.3}^{n + 1}}}}$$

B
$$\frac{{{3^{n + 1}} - 1}}{{{{3}^{n + 1}}}}$$

C
$$\frac{{{3^{n - 1}} - 1}}{{{{3}^{n + 1}}}}$$

D
$$\frac{{{3^{n - 1}} - 1}}{{{{2.3}^{n + 1}}}}$$

Solution
Question 5
1 / -0

The sum of the coefficients in the expansion of $${\left( {1 + x3{x^2}} \right)^{2163}}$$ will be

A
0

B
1

C
-1

D
2

Solution
Question 6
1 / -0

The co-efficient of $$x^5$$ in the expression of $${\left( {1 + x} \right)^{21}} + {\left( {1 + x} \right)^{22}} + .......... + {\left( {1 + x} \right)^{30}}$$ is :

A
$$^{51}{C_5}$$

B
$$^{9}{C_5}$$

C
$$^{31}{C_6}{ - ^{21}}{C_6}$$

D
$$^{30}{C_5}{ - ^{20}}{C_5}$$

Solution
Question 7
1 / -0

The coefficient of $${x^9}$$ in $$\left( {x - 1} \right)\left( {x - 4} \right)\left( {x - 9} \right)......\left( {x - 100} \right)$$ is

A
-235

B
235

C
385

D
-385

Solution
Question 8
1 / -0

If the third term in the binomial expansion of $$(1 + x)^m$$ is $$-\frac{1}{8}x^2$$, the the rational value of m is-

A
$$2$$

B
$$\frac{1}{2}$$

C
$$3$$

D
$$4$$
Question 9
1 / -0

If $${C_0},{C_1},{C_2},.....,{C_n}$$ are the binomial coefficients, then $$2{C_1}+{2^3}{C_3}+{2^5}{C_5} + ...$$ equals

A
$$\dfrac{{{3^n} + {{\left( { - 1} \right)}^n}}}{2}$$

B
$$\dfrac{{{3^n} - {{\left( { - 1} \right)}^n}}}{2}$$

C
$$\dfrac{{{3^n} + 1}}{2}$$

D
$$\dfrac{{{3^n} - 1}}{2}$$

Solution

We have,
$$(1+x)^n =C_0+C_1x +C_2x^2 +C_3x^3+....+C_n x^n\quad...(1)\ \ \ \text{ and }\\(1-x)^n =C_0-C_1x+C_2x^2-C_3x^3+.....+(-1)^n \cdot C_n x^n\quad...(2)$$

Subtracting equation $$(2)$$ from $$(1)$$
$$\Rightarrow (1+x)^n -(1-x)^n =2[C_1x+C_3x^3 +C_5x^5+....]$$
$$\Rightarrow \dfrac{1}{2}[(1+x)^n -(1-x)^n ] =C_1x+C_3x^3+C_5x^5+.....$$

Putting $$x=2$$, we get,
$$2C_1 +2^3 C_3+ 2^5 C_5 +...=\dfrac{3^n -(-1)^n}{2}$$

Hence, the correct answer is option (B).
Question 10
1 / -0

The sum of the coefficient in the expansion of $$(1+5x-7x^2)^{3546}$$ is

A
$$275$$

B
$$1$$

C
$$3546$$

D
$$575$$

Solution